Gọi \(D\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(2;2\right)\\\overrightarrow{DC}=\left(8-x;2-y\right)\end{matrix}\right.\)
\(\overrightarrow{AB}=\overrightarrow{DC}\Rightarrow\left\{{}\begin{matrix}8-x=2\\2-y=2\end{matrix}\right.\) \(\Rightarrow D\left(6;0\right)\)
\(\overrightarrow{DD'}=\overrightarrow{v}=\left(-2;1\right)\Rightarrow D'\left(4;1\right)\) \(\Rightarrow\overrightarrow{D'B}=\left(1;5\right)\)
\(\Rightarrow\) Đường thẳng BD' nhận \(\left(5;-1\right)\) là 1 vtpt
Pt BD': \(5\left(x-5\right)-1\left(y-6\right)=0\Leftrightarrow5x-y-19=0\)
\(\Rightarrow d\left(O;BD'\right)=\frac{\left|-19\right|}{\sqrt{5^2+\left(-1\right)^2}}=\frac{19}{\sqrt{26}}\)
